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Matrix Question Help!!

Can anyone please explain how to go about this c) part? Question is down below. Thank you!!
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Original post by 36 That Boy
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Note that M4M^4 is just −I-I. This means that M8=M4⋅M4=(−I)(−I)=I2=IM^8 = M^4 \cdot M^4 = (-I)(-I) = I^2 = I.

Hence (M8)k=Ik  ⟹  M8k=I(M^8)^k = I^k \implies M^{8k} = I for any integer kk, so you can choose kk so that you get M raised to a very large power being equal to just I. The question asks for M2006M^{2006}, so you might think about getting kk so that 8k=20068k = 2006 ... but unfortunately kk is not an integer if we solve this equation... but is there a number *close* to 2006 which would give you an appropriate kk to choose?
Ummm how about 2008 which when divided by 8 is 251
Original post by RDKGames
Note that M4M^4 is just −I-I. This means that M8=M4⋅M4=(−I)(−I)=I2=IM^8 = M^4 \cdot M^4 = (-I)(-I) = I^2 = I.

Hence (M8)k=Ik  ⟹  M8k=I(M^8)^k = I^k \implies M^{8k} = I for any integer kk, so you can choose kk so that you get M raised to a very large power being equal to just I. The question asks for M2006M^{2006}, so you might think about getting kk so that 8k=20068k = 2006 ... but unfortunately kk is not an integer if we solve this equation... but is there a number *close* to 2006 which would give you an appropriate kk to choose?
Original post by 36 That Boy
Ummm how about 2008 which when divided by 8 is 251


Sounds good, but since you're not confident on this question yet, I suggest you pick k=250k = 250 as to then obtain the power of 20002000 [since]

So then, M2000=IM^{2000} = I.

But you're interested in M2006M^{2006}. What can you do to the LHS of our equation in order to go from M2000M^{2000} to M2006M^{2006} ?? Do the same to the RHS. What do you get?
Multiply by M^6 on both sides? Then i get M^2006 = M^6 which is M^4 * M^2 calculated previously?
Original post by RDKGames
Sounds good, but since you're not confident on this question yet, I suggest you pick k=250k = 250 as to then obtain the power of 20002000 [since]

So then, M2000=IM^{2000} = I.

But you're interested in M2006M^{2006}. What can you do to the LHS of our equation in order to go from M2000M^{2000} to M2006M^{2006} ?? Do the same to the RHS. What do you get?
Original post by 36 That Boy
Multiply by M^6 on both sides? Then i get M^2006 = M^6 which is M^4 * M^2 calculated previously?


That's right, and using the fact that M^4 = -I you shouldn't need to multiply any matrices out.
Is there any way to know that M^4 = -I without calculating M^4?
Original post by RDKGames
That's right, and using the fact that M^4 = -I you shouldn't need to multiply any matrices out.
Original post by 36 That Boy
Is there any way to know that M^4 = -I without calculating M^4?


It's not so obvious without calculating it. Of course, in this question you're asked to calculate it in the previous part, so you know what it is. The rest is just to realise that is looks very similar to the identity matrix... indeed it's just the -ve of it.

There is a way without any calculation, but it hinges on your understanding of matrix transformations visually. For instance, this matrix M represents a rotation clockwise about the origin by 45 degrees. M^4 means you apply this 4 times, so it's a 45 x 4 = 180 degree rotation. Where do the unit vectors (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} and (01)\begin{pmatrix} 0 \\ 1 \end{pmatrix} go after this transformation? Obviously to (−10)\begin{pmatrix} -1 \\ 0 \end{pmatrix} and (0−1)\begin{pmatrix} 0 \\ -1 \end{pmatrix} respectively... so the matrix for M^4 is (−100−1)\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}.
(edited 4 years ago)
This makes so much sense now. But doesn’t M represent 45° rotation clockwise about the origin since It's in the 4th quadrant?
Original post by RDKGames
It's not so obvious without calculating it. Of course, in this question you're asked to calculate it in the previous part, so you know what it is. The rest is just to realise that is looks very similar to the identity matrix... indeed it's just the -ve of it.

There is a way without any calculation, but it hinges on your understanding of matrix transformations visually. For instance, this matrix M represents a rotation anticlockwise about the origin by 45 degrees. M^4 means you apply this 4 times, so it's a 45 x 4 = 180 degree rotation. Where do the unit vectors (10)\begin{pmatrix} 1 \\ 0 \end{pmatrix} and (01)\begin{pmatrix} 0 \\ 1 \end{pmatrix} go after this transformation? Obviously to (−10)\begin{pmatrix} -1 \\ 0 \end{pmatrix} and (0−1)\begin{pmatrix} 0 \\ -1 \end{pmatrix} respectively... so the matrix for M^4 is (−100−1)\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}.
(edited 4 years ago)
Original post by 36 That Boy
This makes so much sense now. But doesn’t M represent 45° rotation clockwise about the origin since It's in the 4th quadrant?


Yeah, that's what I meant :smile: it's 4 lots of it at the end of the day so it doesn't matter whether we go clockwise or anticlockwise for M^4
Thanks a lot for your help.Out of curiosity what if i wanted M^2007? I tried it and it doesn't seem to get me anywhere cause 2007 is an odd number?
Original post by RDKGames
Yeah, that's what I meant :smile: it's 4 lots of it at the end of the day so it doesn't matter whether we go clockwise or anticlockwise for M^4
Original post by 36 That Boy
Thanks a lot for your help.Out of curiosity what if i wanted M^2007? I tried it and it doesn't seem to get me anywhere cause 2007 is an odd number?


Just multiply M2000=IM^{2000} = I through by M7=M4⋅M2⋅MM^7 = M^{4} \cdot M^{2} \cdot M.


Alternatively, if you're OK with taking inverse matrices, then clearly we have established by that M2008=IM^{2008} = I, so multiplying through by M−1M^{-1} will also yield the result.
Ah i see. That's just perfect. Thanks for your priceless help mate 😊✌
Original post by RDKGames
Just multiply M2000=IM^{2000} = I through by M7=M4⋅M2⋅MM^7 = M^{4} \cdot M^{2} \cdot M.


Alternatively, if you're OK with taking inverse matrices, then clearly we have established by that M2008=IM^{2008} = I, so multiplying through by M−1M^{-1} will also yield the result.

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